d) \( a^{x}+b^{y}=(a b)^{x y} \)

To analyze the equation \( a^{x} + b^{y} = (ab)^{xy} \), we can start by rewriting the right-hand side: \[ (ab)^{xy} = a^{xy} b^{xy} \] Thus, the equation can be rewritten as: \[ a^{x} + b^{y} = a^{xy} b^{xy} \] This equation suggests a relationship between the terms \( a^x \), \( b^y \), and \( a^{xy} b^{xy} \). To explore this further, we can consider specific values for \( a \), \( b \), \( x \), and \( y \). ### Case 1: \( a = b = 1 \) If we set \( a = 1 \) and \( b = 1 \): \[ 1^{x} + 1^{y} = (1 \cdot 1)^{xy} \implies 1 + 1 = 1 \implies 2 = 1 \] This is not true, so \( a = 1 \) and \( b = 1 \) does not satisfy the equation. ### Case 2: \( a = b = 2 \) If we set \( a = 2 \) and \( b = 2 \): \[ 2^{x} + 2^{y} = (2 \cdot 2)^{xy} \implies 2^{x} + 2^{y} = 4^{xy} \implies 2^{x} + 2^{y} = (2^2)^{xy} = 2^{2xy} \] This simplifies to: \[ 2^{x} + 2^{y} = 2^{2xy} \] This equation holds true under certain conditions for \( x \) and \( y \). For example, if \( x = y = 1 \): \[ 2^{1} + 2^{1} = 2^{2} \implies 2 + 2 = 4 \implies 4 = 4 \] This is true. ### General Case To find a more general solution, we can analyze the equation further. 1. **Assuming \( x = y \)**: Let \( x = y = k \): \[ a^{k} + b^{k} = (ab)^{k^2} \implies a^{k} + b^{k} = a^{k^2} b^{k^2} \] This can lead to specific solutions depending on the values of \( a \) and \( b \). 2. **Exploring limits**: As \( x \) and \( y \) grow large, the left-hand side \( a^x + b^y \) will be dominated by the larger of the two terms, while the right-hand side grows as \( (ab)^{xy} \). ### Conclusion The equation \( a^{x} + b^{y} = (ab)^{xy} \) does not hold for all values of \( a \), \( b \), \( x \), and \( y \). Specific cases can yield valid solutions, particularly when \( x \) and \( y \) are equal or when \( a \) and \( b \) are equal. Further exploration into the nature of \( a \) and \( b \) (e.g., whether they are greater than 1, equal, etc.) can yield more insights into the conditions under which the equation holds true.